Question 1208807
{{{f(x)=2(x^2-7x/2-5)}}}

{{{2(x^2-7x/2+(7/4)^2-(7/4)^2-5)}}}

{{{2((x-7/4)^2-49/16-5)}}}

{{{2(x-7/4)^2-49/8-10}}}

{{{2(x-7/4)^2-129/8}}}

{{{highlight(2(x-7/4)^2-16&1/8)}}}

The vertex here will be the lowest point of the graph
which is at  (1&3/4, -16&1/8).  
{{{h=7/4}}} and {{{k=-16&1/8}}}.


If solve f(x) equated at 0, then you find the roots.
 {{{2(x-7/4)^2-16&1/8=0}}}.


{{{2(x-7/4)^2-129/8=0}}}

{{{2(x-7/4)^2=129/8}}}

{{{(x-7/4)^2=129/16}}}

{{{x-7/4=0+- sqrt(129)/4}}}

{{{highlight(x=7/4+- sqrt(129)/4)}}} the roots